using System;
using System.Diagnostics;
using Jint.Runtime;

namespace Jint.Native.Number.Dtoa
{
    internal static class BignumDtoa
    {
        public static void NumberToString(
            double v,
            DtoaMode mode,
            int requested_digits,
            DtoaBuilder builder,
            out int decimal_point)
        {
            var bits = (ulong) BitConverter.DoubleToInt64Bits(v);
            var significand = DoubleHelper.Significand(bits);
            var is_even = (significand & 1) == 0;
            var exponent = DoubleHelper.Exponent(bits);
            var normalized_exponent = DoubleHelper.NormalizedExponent(significand, exponent);
            // estimated_power might be too low by 1.
            var estimated_power = EstimatePower(normalized_exponent);

            // Shortcut for Fixed.
            // The requested digits correspond to the digits after the point. If the
            // number is much too small, then there is no need in trying to get any
            // digits.
            if (mode == DtoaMode.Fixed && -estimated_power - 1 > requested_digits)
            {
                // Set decimal-point to -requested_digits. This is what Gay does.
                // Note that it should not have any effect anyways since the string is
                // empty.
                decimal_point = -requested_digits;
                return;
            }

            Bignum numerator = new Bignum();
            Bignum denominator = new Bignum();
            Bignum delta_minus = new Bignum();
            Bignum delta_plus = new Bignum();
            // Make sure the bignum can grow large enough. The smallest double equals
            // 4e-324. In this case the denominator needs fewer than 324*4 binary digits.
            // The maximum double is 1.7976931348623157e308 which needs fewer than
            // 308*4 binary digits.
            var need_boundary_deltas = mode == DtoaMode.Shortest;

            InitialScaledStartValues(
                v,
                estimated_power,
                need_boundary_deltas,
                numerator,
                denominator,
                delta_minus,
                delta_plus);
            // We now have v = (numerator / denominator) * 10^estimated_power.
            FixupMultiply10(
                estimated_power,
                is_even,
                out decimal_point,
                numerator,
                denominator,
                delta_minus,
                delta_plus);
            // We now have v = (numerator / denominator) * 10^(decimal_point-1), and
            //  1 <= (numerator + delta_plus) / denominator < 10
            switch (mode)
            {
                case DtoaMode.Shortest:
                    GenerateShortestDigits(
                        numerator,
                        denominator,
                        delta_minus,
                        delta_plus,
                        is_even,
                        builder);
                    break;
                case DtoaMode.Fixed:
                    BignumToFixed(
                        requested_digits,
                        ref decimal_point,
                        numerator,
                        denominator,
                        builder);
                    break;
                case DtoaMode.Precision:
                    GenerateCountedDigits(
                        requested_digits,
                        ref decimal_point,
                        numerator,
                        denominator,
                        builder);
                    break;
                default:
                    ExceptionHelper.ThrowArgumentOutOfRangeException();
                    break;
            }
        }


        // The procedure starts generating digits from the left to the right and stops
        // when the generated digits yield the shortest decimal representation of v. A
        // decimal representation of v is a number lying closer to v than to any other
        // double, so it converts to v when read.
        //
        // This is true if d, the decimal representation, is between m- and m+, the
        // upper and lower boundaries. d must be strictly between them if !is_even.
        //           m- := (numerator - delta_minus) / denominator
        //           m+ := (numerator + delta_plus) / denominator
        //
        // Precondition: 0 <= (numerator+delta_plus) / denominator < 10.
        //   If 1 <= (numerator+delta_plus) / denominator < 10 then no leading 0 digit
        //   will be produced. This should be the standard precondition.
        private static void GenerateShortestDigits(
            Bignum numerator,
            Bignum denominator,
            Bignum delta_minus,
            Bignum delta_plus,
            bool is_even,
            DtoaBuilder buffer)
        {
            // Small optimization: if delta_minus and delta_plus are the same just reuse
            // one of the two bignums.
            if (Bignum.Equal(delta_minus, delta_plus))
            {
                delta_plus = delta_minus;
            }

            buffer.Reset();
            while (true)
            {
                uint digit;
                digit = numerator.DivideModuloIntBignum(denominator);
                Debug.Assert(digit <= 9); // digit is a uint and therefore always positive.
                // digit = numerator / denominator (integer division).
                // numerator = numerator % denominator.
                buffer.Append((char) (digit + '0'));

                // Can we stop already?
                // If the remainder of the division is less than the distance to the lower
                // boundary we can stop. In this case we simply round down (discarding the
                // remainder).
                // Similarly we test if we can round up (using the upper boundary).
                bool in_delta_room_minus;
                bool in_delta_room_plus;
                if (is_even)
                {
                    in_delta_room_minus = Bignum.LessEqual(numerator, delta_minus);
                }
                else
                {
                    in_delta_room_minus = Bignum.Less(numerator, delta_minus);
                }
                if (is_even)
                {
                    in_delta_room_plus = Bignum.PlusCompare(numerator, delta_plus, denominator) >= 0;
                }
                else
                {
                    in_delta_room_plus = Bignum.PlusCompare(numerator, delta_plus, denominator) > 0;
                }
                if (!in_delta_room_minus && !in_delta_room_plus)
                {
                    // Prepare for next iteration.
                    numerator.Times10();
                    delta_minus.Times10();
                    // We optimized delta_plus to be equal to delta_minus (if they share the
                    // same value). So don't multiply delta_plus if they point to the same
                    // object.
                    if (delta_minus != delta_plus) delta_plus.Times10();
                }
                else if (in_delta_room_minus && in_delta_room_plus)
                {
                    // Let's see if 2*numerator < denominator.
                    // If yes, then the next digit would be < 5 and we can round down.
                    int compare = Bignum.PlusCompare(numerator, numerator, denominator);
                    if (compare < 0)
                    {
                        // Remaining digits are less than .5. -> Round down (== do nothing).
                    }
                    else if (compare > 0)
                    {
                        // Remaining digits are more than .5 of denominator. . Round up.
                        // Note that the last digit could not be a '9' as otherwise the whole
                        // loop would have stopped earlier.
                        // We still have an assert here in case the preconditions were not
                        // satisfied.
                        Debug.Assert(buffer[buffer.Length - 1] != '9');
                        buffer[buffer.Length - 1]++;
                    }
                    else
                    {
                        // Halfway case.
                        // TODO(floitsch): need a way to solve half-way cases.
                        //   For now let's round towards even (since this is what Gay seems to
                        //   do).

                        if ((buffer[buffer.Length - 1] - '0') % 2 == 0)
                        {
                            // Round down => Do nothing.
                        }
                        else
                        {
                            Debug.Assert(buffer[buffer.Length - 1] != '9');
                            buffer[buffer.Length - 1]++;
                        }
                    }

                    return;
                }
                else if (in_delta_room_minus)
                {
                    // Round down (== do nothing).
                    return;
                }
                else
                {
                    // in_delta_room_plus
                    // Round up.
                    // Note again that the last digit could not be '9' since this would have
                    // stopped the loop earlier.
                    // We still have an DCHECK here, in case the preconditions were not
                    // satisfied.
                    Debug.Assert(buffer[buffer.Length - 1] != '9');
                    buffer[buffer.Length - 1]++;
                    return;
                }
            }
        }
        
        // Let v = numerator / denominator < 10.
        // Then we generate 'count' digits of d = x.xxxxx... (without the decimal point)
        // from left to right. Once 'count' digits have been produced we decide wether
        // to round up or down. Remainders of exactly .5 round upwards. Numbers such
        // as 9.999999 propagate a carry all the way, and change the
        // exponent (decimal_point), when rounding upwards.
        static void GenerateCountedDigits(
            int count, 
            ref int decimal_point,
            Bignum numerator, 
            Bignum denominator,
            DtoaBuilder buffer)
        {
            Debug.Assert(count >= 0);
            for (int i = 0; i < count - 1; ++i)
            {
                uint d = numerator.DivideModuloIntBignum(denominator);
                Debug.Assert(d <= 9);  // digit is a uint and therefore always positive.
                // digit = numerator / denominator (integer division).
                // numerator = numerator % denominator.
                buffer.Append((char) (d + '0'));
                // Prepare for next iteration.
                numerator.Times10();
            }
            // Generate the last digit.
            uint digit = numerator.DivideModuloIntBignum(denominator);
            if (Bignum.PlusCompare(numerator, numerator, denominator) >= 0)
            {
                digit++;
            }
            buffer.Append((char) (digit + '0'));
            // Correct bad digits (in case we had a sequence of '9's). Propagate the
            // carry until we hat a non-'9' or til we reach the first digit.
            for (int i = count - 1; i > 0; --i)
            {
                if (buffer[i] != '0' + 10) break;
                buffer[i] = '0';
                buffer[i - 1]++;
            }
            if (buffer[0] == '0' + 10)
            {
                // Propagate a carry past the top place.
                buffer[0] = '1';
                decimal_point++;
            }
        }


        // Generates 'requested_digits' after the decimal point. It might omit
        // trailing '0's. If the input number is too small then no digits at all are
        // generated (ex.: 2 fixed digits for 0.00001).
        //
        // Input verifies:  1 <= (numerator + delta) / denominator < 10.
        static void BignumToFixed(
            int requested_digits,
            ref int decimal_point,
            Bignum numerator,
            Bignum denominator,
            DtoaBuilder buffer)
        {
            // Note that we have to look at more than just the requested_digits, since
            // a number could be rounded up. Example: v=0.5 with requested_digits=0.
            // Even though the power of v equals 0 we can't just stop here.
            if (-(decimal_point) > requested_digits)
            {
                // The number is definitively too small.
                // Ex: 0.001 with requested_digits == 1.
                // Set decimal-point to -requested_digits. This is what Gay does.
                // Note that it should not have any effect anyways since the string is
                // empty.
                decimal_point = -requested_digits;
                buffer.Reset();
                return;
            }

            if (-decimal_point == requested_digits)
            {
                // We only need to verify if the number rounds down or up.
                // Ex: 0.04 and 0.06 with requested_digits == 1.
                Debug.Assert(decimal_point == -requested_digits);
                // Initially the fraction lies in range (1, 10]. Multiply the denominator
                // by 10 so that we can compare more easily.
                denominator.Times10();
                if (Bignum.PlusCompare(numerator, numerator, denominator) >= 0)
                {
                    // If the fraction is >= 0.5 then we have to include the rounded
                    // digit.
                    buffer[0] = '1';
                    decimal_point++;
                }
                else
                {
                    // Note that we caught most of similar cases earlier.
                    buffer.Reset();
                }
            }
            else
            {
                // The requested digits correspond to the digits after the point.
                // The variable 'needed_digits' includes the digits before the point.
                int needed_digits = (decimal_point) + requested_digits;
                GenerateCountedDigits(needed_digits, ref decimal_point, numerator, denominator, buffer);
            }
        }

        // Returns an estimation of k such that 10^(k-1) <= v < 10^k where
        // v = f * 2^exponent and 2^52 <= f < 2^53.
        // v is hence a normalized double with the given exponent. The output is an
        // approximation for the exponent of the decimal approimation .digits * 10^k.
        //
        // The result might undershoot by 1 in which case 10^k <= v < 10^k+1.
        // Note: this property holds for v's upper boundary m+ too.
        //    10^k <= m+ < 10^k+1.
        //   (see explanation below).
        //
        // Examples:
        //  EstimatePower(0)   => 16
        //  EstimatePower(-52) => 0
        //
        // Note: e >= 0 => EstimatedPower(e) > 0. No similar claim can be made for e<0.
        private static int EstimatePower(int exponent)
        {
            // This function estimates log10 of v where v = f*2^e (with e == exponent).
            // Note that 10^floor(log10(v)) <= v, but v <= 10^ceil(log10(v)).
            // Note that f is bounded by its container size. Let p = 53 (the double's
            // significand size). Then 2^(p-1) <= f < 2^p.
            //
            // Given that log10(v) == log2(v)/log2(10) and e+(len(f)-1) is quite close
            // to log2(v) the function is simplified to (e+(len(f)-1)/log2(10)).
            // The computed number undershoots by less than 0.631 (when we compute log3
            // and not log10).
            //
            // Optimization: since we only need an approximated result this computation
            // can be performed on 64 bit integers. On x86/x64 architecture the speedup is
            // not really measurable, though.
            //
            // Since we want to avoid overshooting we decrement by 1e10 so that
            // floating-point imprecisions don't affect us.
            //
            // Explanation for v's boundary m+: the computation takes advantage of
            // the fact that 2^(p-1) <= f < 2^p. Boundaries still satisfy this requirement
            // (even for denormals where the delta can be much more important).

            const double k1Log10 = 0.30102999566398114; // 1/lg(10)

            // For doubles len(f) == 53 (don't forget the hidden bit).
            const int kSignificandSize = 53;
            double estimate = System.Math.Ceiling((exponent + kSignificandSize - 1) * k1Log10 - 1e-10);
            return (int) estimate;
        }


        // See comments for InitialScaledStartValues.
        private static void InitialScaledStartValuesPositiveExponent(
            double v,
            int estimated_power,
            bool need_boundary_deltas,
            Bignum numerator,
            Bignum denominator,
            Bignum delta_minus,
            Bignum delta_plus)
        {
            // A positive exponent implies a positive power.
            Debug.Assert(estimated_power >= 0);
            // Since the estimated_power is positive we simply multiply the denominator
            // by 10^estimated_power.

            // numerator = v.
            var bits = (ulong) BitConverter.DoubleToInt64Bits(v);
            numerator.AssignUInt64(DoubleHelper.Significand(bits));
            numerator.ShiftLeft(DoubleHelper.Exponent(bits));
            // denominator = 10^estimated_power.
            denominator.AssignPowerUInt16(10, estimated_power);

            if (need_boundary_deltas)
            {
                // Introduce a common denominator so that the deltas to the boundaries are
                // integers.
                denominator.ShiftLeft(1);
                numerator.ShiftLeft(1);
                // Let v = f * 2^e, then m+ - v = 1/2 * 2^e; With the common
                // denominator (of 2) delta_plus equals 2^e.
                delta_plus.AssignUInt16(1);
                delta_plus.ShiftLeft(DoubleHelper.Exponent(bits));
                // Same for delta_minus (with adjustments below if f == 2^p-1).
                delta_minus.AssignUInt16(1);
                delta_minus.ShiftLeft(DoubleHelper.Exponent(bits));

                // If the significand (without the hidden bit) is 0, then the lower
                // boundary is closer than just half a ulp (unit in the last place).
                // There is only one exception: if the next lower number is a denormal then
                // the distance is 1 ulp. This cannot be the case for exponent >= 0 (but we
                // have to test it in the other function where exponent < 0).
                ulong v_bits = bits;
                if ((v_bits & DoubleHelper.KSignificandMask) == 0)
                {
                    // The lower boundary is closer at half the distance of "normal" numbers.
                    // Increase the common denominator and adapt all but the delta_minus.
                    denominator.ShiftLeft(1); // *2
                    numerator.ShiftLeft(1); // *2
                    delta_plus.ShiftLeft(1); // *2
                }
            }
        }


        // See comments for InitialScaledStartValues
        private static void InitialScaledStartValuesNegativeExponentPositivePower(
            double v,
            int estimated_power, 
            bool need_boundary_deltas,
            Bignum numerator, 
            Bignum denominator,
            Bignum delta_minus, 
            Bignum delta_plus)
        {
            var bits = (ulong) BitConverter.DoubleToInt64Bits(v);
            ulong significand = DoubleHelper.Significand(bits);
            int exponent = DoubleHelper.Exponent(bits);
            // v = f * 2^e with e < 0, and with estimated_power >= 0.
            // This means that e is close to 0 (have a look at how estimated_power is
            // computed).

            // numerator = significand
            //  since v = significand * 2^exponent this is equivalent to
            //  numerator = v * / 2^-exponent
            numerator.AssignUInt64(significand);
            // denominator = 10^estimated_power * 2^-exponent (with exponent < 0)
            denominator.AssignPowerUInt16(10, estimated_power);
            denominator.ShiftLeft(-exponent);

            if (need_boundary_deltas)
            {
                // Introduce a common denominator so that the deltas to the boundaries are
                // integers.
                denominator.ShiftLeft(1);
                numerator.ShiftLeft(1);
                // Let v = f * 2^e, then m+ - v = 1/2 * 2^e; With the common
                // denominator (of 2) delta_plus equals 2^e.
                // Given that the denominator already includes v's exponent the distance
                // to the boundaries is simply 1.
                delta_plus.AssignUInt16(1);
                // Same for delta_minus (with adjustments below if f == 2^p-1).
                delta_minus.AssignUInt16(1);

                // If the significand (without the hidden bit) is 0, then the lower
                // boundary is closer than just one ulp (unit in the last place).
                // There is only one exception: if the next lower number is a denormal
                // then the distance is 1 ulp. Since the exponent is close to zero
                // (otherwise estimated_power would have been negative) this cannot happen
                // here either.
                ulong v_bits = bits;
                if ((v_bits & DoubleHelper.KSignificandMask) == 0)
                {
                    // The lower boundary is closer at half the distance of "normal" numbers.
                    // Increase the denominator and adapt all but the delta_minus.
                    denominator.ShiftLeft(1); // *2
                    numerator.ShiftLeft(1); // *2
                    delta_plus.ShiftLeft(1); // *2
                }
            }
        }


        // See comments for InitialScaledStartValues
        private static void InitialScaledStartValuesNegativeExponentNegativePower(
            double v,
            int estimated_power,
            bool need_boundary_deltas,
            Bignum numerator,
            Bignum denominator,
            Bignum delta_minus,
            Bignum delta_plus)
        {
            const ulong kMinimalNormalizedExponent = 0x0010000000000000;

            var bits = (ulong) BitConverter.DoubleToInt64Bits(v);
            ulong significand = DoubleHelper.Significand(bits);
            int exponent = DoubleHelper.Exponent(bits);
            // Instead of multiplying the denominator with 10^estimated_power we
            // multiply all values (numerator and deltas) by 10^-estimated_power.

            // Use numerator as temporary container for power_ten.
            Bignum power_ten = numerator;
            power_ten.AssignPowerUInt16(10, -estimated_power);

            if (need_boundary_deltas)
            {
                // Since power_ten == numerator we must make a copy of 10^estimated_power
                // before we complete the computation of the numerator.
                // delta_plus = delta_minus = 10^estimated_power
                delta_plus.AssignBignum(power_ten);
                delta_minus.AssignBignum(power_ten);
            }

            // numerator = significand * 2 * 10^-estimated_power
            //  since v = significand * 2^exponent this is equivalent to
            // numerator = v * 10^-estimated_power * 2 * 2^-exponent.
            // Remember: numerator has been abused as power_ten. So no need to assign it
            //  to itself.
            Debug.Assert(numerator == power_ten);
            numerator.MultiplyByUInt64(significand);

            // denominator = 2 * 2^-exponent with exponent < 0.
            denominator.AssignUInt16(1);
            denominator.ShiftLeft(-exponent);

            if (need_boundary_deltas)
            {
                // Introduce a common denominator so that the deltas to the boundaries are
                // integers.
                numerator.ShiftLeft(1);
                denominator.ShiftLeft(1);
                // With this shift the boundaries have their correct value, since
                // delta_plus = 10^-estimated_power, and
                // delta_minus = 10^-estimated_power.
                // These assignments have been done earlier.

                // The special case where the lower boundary is twice as close.
                // This time we have to look out for the exception too.
                ulong v_bits = bits;
                if ((v_bits & DoubleHelper.KSignificandMask) == 0 &&
                    // The only exception where a significand == 0 has its boundaries at
                    // "normal" distances:
                    (v_bits & DoubleHelper.KExponentMask) != kMinimalNormalizedExponent)
                {
                    numerator.ShiftLeft(1); // *2
                    denominator.ShiftLeft(1); // *2
                    delta_plus.ShiftLeft(1); // *2
                }
            }
        }


        // Let v = significand * 2^exponent.
        // Computes v / 10^estimated_power exactly, as a ratio of two bignums, numerator
        // and denominator. The functions GenerateShortestDigits and
        // GenerateCountedDigits will then convert this ratio to its decimal
        // representation d, with the required accuracy.
        // Then d * 10^estimated_power is the representation of v.
        // (Note: the fraction and the estimated_power might get adjusted before
        // generating the decimal representation.)
        //
        // The initial start values consist of:
        //  - a scaled numerator: s.t. numerator/denominator == v / 10^estimated_power.
        //  - a scaled (common) denominator.
        //  optionally (used by GenerateShortestDigits to decide if it has the shortest
        //  decimal converting back to v):
        //  - v - m-: the distance to the lower boundary.
        //  - m+ - v: the distance to the upper boundary.
        //
        // v, m+, m-, and therefore v - m- and m+ - v all share the same denominator.
        //
        // Let ep == estimated_power, then the returned values will satisfy:
        //  v / 10^ep = numerator / denominator.
        //  v's boundarys m- and m+:
        //    m- / 10^ep == v / 10^ep - delta_minus / denominator
        //    m+ / 10^ep == v / 10^ep + delta_plus / denominator
        //  Or in other words:
        //    m- == v - delta_minus * 10^ep / denominator;
        //    m+ == v + delta_plus * 10^ep / denominator;
        //
        // Since 10^(k-1) <= v < 10^k    (with k == estimated_power)
        //  or       10^k <= v < 10^(k+1)
        //  we then have 0.1 <= numerator/denominator < 1
        //           or    1 <= numerator/denominator < 10
        //
        // It is then easy to kickstart the digit-generation routine.
        //
        // The boundary-deltas are only filled if need_boundary_deltas is set.
        private static void InitialScaledStartValues(
            double v,
            int estimated_power,
            bool need_boundary_deltas,
            Bignum numerator,
            Bignum denominator,
            Bignum delta_minus,
            Bignum delta_plus)
        {
            var bits = (ulong) BitConverter.DoubleToInt64Bits(v);
            if (DoubleHelper.Exponent(bits) >= 0)
            {
                InitialScaledStartValuesPositiveExponent(
                    v,
                    estimated_power,
                    need_boundary_deltas,
                    numerator,
                    denominator,
                    delta_minus,
                    delta_plus);
            }
            else if (estimated_power >= 0)
            {
                InitialScaledStartValuesNegativeExponentPositivePower(
                    v,
                    estimated_power,
                    need_boundary_deltas,
                    numerator,
                    denominator,
                    delta_minus,
                    delta_plus);
            }
            else
            {
                InitialScaledStartValuesNegativeExponentNegativePower(
                    v,
                    estimated_power,
                    need_boundary_deltas,
                    numerator,
                    denominator,
                    delta_minus,
                    delta_plus);
            }
        }


        // This routine multiplies numerator/denominator so that its values lies in the
        // range 1-10. That is after a call to this function we have:
        //    1 <= (numerator + delta_plus) /denominator < 10.
        // Let numerator the input before modification and numerator' the argument
        // after modification, then the output-parameter decimal_point is such that
        //  numerator / denominator * 10^estimated_power ==
        //    numerator' / denominator' * 10^(decimal_point - 1)
        // In some cases estimated_power was too low, and this is already the case. We
        // then simply adjust the power so that 10^(k-1) <= v < 10^k (with k ==
        // estimated_power) but do not touch the numerator or denominator.
        // Otherwise the routine multiplies the numerator and the deltas by 10.
        private static void FixupMultiply10(
            int estimated_power,
            bool is_even,
            out int decimal_point,
            Bignum numerator,
            Bignum denominator,
            Bignum delta_minus,
            Bignum delta_plus)
        {
            bool in_range;
            if (is_even)
                in_range = Bignum.PlusCompare(numerator, delta_plus, denominator) >= 0;
            else
                in_range = Bignum.PlusCompare(numerator, delta_plus, denominator) > 0;
            if (in_range)
            {
                // Since numerator + delta_plus >= denominator we already have
                // 1 <= numerator/denominator < 10. Simply update the estimated_power.
                decimal_point = estimated_power + 1;
            }
            else
            {
                decimal_point = estimated_power;
                numerator.Times10();
                if (Bignum.Equal(delta_minus, delta_plus))
                {
                    delta_minus.Times10();
                    delta_plus.AssignBignum(delta_minus);
                }
                else
                {
                    delta_minus.Times10();
                    delta_plus.Times10();
                }
            }
        }
    }
}